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University Physics Volume 1

4.2 Acceleration Vector

University Physics Volume 14.2 Acceleration Vector
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  1. Preface
  2. Unit 1. Mechanics
    1. 1 Units and Measurement
      1. Introduction
      2. 1.1 The Scope and Scale of Physics
      3. 1.2 Units and Standards
      4. 1.3 Unit Conversion
      5. 1.4 Dimensional Analysis
      6. 1.5 Estimates and Fermi Calculations
      7. 1.6 Significant Figures
      8. 1.7 Solving Problems in Physics
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 2 Vectors
      1. Introduction
      2. 2.1 Scalars and Vectors
      3. 2.2 Coordinate Systems and Components of a Vector
      4. 2.3 Algebra of Vectors
      5. 2.4 Products of Vectors
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 3 Motion Along a Straight Line
      1. Introduction
      2. 3.1 Position, Displacement, and Average Velocity
      3. 3.2 Instantaneous Velocity and Speed
      4. 3.3 Average and Instantaneous Acceleration
      5. 3.4 Motion with Constant Acceleration
      6. 3.5 Free Fall
      7. 3.6 Finding Velocity and Displacement from Acceleration
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    4. 4 Motion in Two and Three Dimensions
      1. Introduction
      2. 4.1 Displacement and Velocity Vectors
      3. 4.2 Acceleration Vector
      4. 4.3 Projectile Motion
      5. 4.4 Uniform Circular Motion
      6. 4.5 Relative Motion in One and Two Dimensions
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    5. 5 Newton's Laws of Motion
      1. Introduction
      2. 5.1 Forces
      3. 5.2 Newton's First Law
      4. 5.3 Newton's Second Law
      5. 5.4 Mass and Weight
      6. 5.5 Newton’s Third Law
      7. 5.6 Common Forces
      8. 5.7 Drawing Free-Body Diagrams
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    6. 6 Applications of Newton's Laws
      1. Introduction
      2. 6.1 Solving Problems with Newton’s Laws
      3. 6.2 Friction
      4. 6.3 Centripetal Force
      5. 6.4 Drag Force and Terminal Speed
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    7. 7 Work and Kinetic Energy
      1. Introduction
      2. 7.1 Work
      3. 7.2 Kinetic Energy
      4. 7.3 Work-Energy Theorem
      5. 7.4 Power
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    8. 8 Potential Energy and Conservation of Energy
      1. Introduction
      2. 8.1 Potential Energy of a System
      3. 8.2 Conservative and Non-Conservative Forces
      4. 8.3 Conservation of Energy
      5. 8.4 Potential Energy Diagrams and Stability
      6. 8.5 Sources of Energy
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
    9. 9 Linear Momentum and Collisions
      1. Introduction
      2. 9.1 Linear Momentum
      3. 9.2 Impulse and Collisions
      4. 9.3 Conservation of Linear Momentum
      5. 9.4 Types of Collisions
      6. 9.5 Collisions in Multiple Dimensions
      7. 9.6 Center of Mass
      8. 9.7 Rocket Propulsion
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    10. 10 Fixed-Axis Rotation
      1. Introduction
      2. 10.1 Rotational Variables
      3. 10.2 Rotation with Constant Angular Acceleration
      4. 10.3 Relating Angular and Translational Quantities
      5. 10.4 Moment of Inertia and Rotational Kinetic Energy
      6. 10.5 Calculating Moments of Inertia
      7. 10.6 Torque
      8. 10.7 Newton’s Second Law for Rotation
      9. 10.8 Work and Power for Rotational Motion
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    11. 11 Angular Momentum
      1. Introduction
      2. 11.1 Rolling Motion
      3. 11.2 Angular Momentum
      4. 11.3 Conservation of Angular Momentum
      5. 11.4 Precession of a Gyroscope
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    12. 12 Static Equilibrium and Elasticity
      1. Introduction
      2. 12.1 Conditions for Static Equilibrium
      3. 12.2 Examples of Static Equilibrium
      4. 12.3 Stress, Strain, and Elastic Modulus
      5. 12.4 Elasticity and Plasticity
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    13. 13 Gravitation
      1. Introduction
      2. 13.1 Newton's Law of Universal Gravitation
      3. 13.2 Gravitation Near Earth's Surface
      4. 13.3 Gravitational Potential Energy and Total Energy
      5. 13.4 Satellite Orbits and Energy
      6. 13.5 Kepler's Laws of Planetary Motion
      7. 13.6 Tidal Forces
      8. 13.7 Einstein's Theory of Gravity
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    14. 14 Fluid Mechanics
      1. Introduction
      2. 14.1 Fluids, Density, and Pressure
      3. 14.2 Measuring Pressure
      4. 14.3 Pascal's Principle and Hydraulics
      5. 14.4 Archimedes’ Principle and Buoyancy
      6. 14.5 Fluid Dynamics
      7. 14.6 Bernoulli’s Equation
      8. 14.7 Viscosity and Turbulence
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  3. Unit 2. Waves and Acoustics
    1. 15 Oscillations
      1. Introduction
      2. 15.1 Simple Harmonic Motion
      3. 15.2 Energy in Simple Harmonic Motion
      4. 15.3 Comparing Simple Harmonic Motion and Circular Motion
      5. 15.4 Pendulums
      6. 15.5 Damped Oscillations
      7. 15.6 Forced Oscillations
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 16 Waves
      1. Introduction
      2. 16.1 Traveling Waves
      3. 16.2 Mathematics of Waves
      4. 16.3 Wave Speed on a Stretched String
      5. 16.4 Energy and Power of a Wave
      6. 16.5 Interference of Waves
      7. 16.6 Standing Waves and Resonance
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 17 Sound
      1. Introduction
      2. 17.1 Sound Waves
      3. 17.2 Speed of Sound
      4. 17.3 Sound Intensity
      5. 17.4 Normal Modes of a Standing Sound Wave
      6. 17.5 Sources of Musical Sound
      7. 17.6 Beats
      8. 17.7 The Doppler Effect
      9. 17.8 Shock Waves
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  4. A | Units
  5. B | Conversion Factors
  6. C | Fundamental Constants
  7. D | Astronomical Data
  8. E | Mathematical Formulas
  9. F | Chemistry
  10. G | The Greek Alphabet
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
    13. Chapter 13
    14. Chapter 14
    15. Chapter 15
    16. Chapter 16
    17. Chapter 17
  12. Index

Learning Objectives

By the end of this section, you will be able to:
  • Calculate the acceleration vector given the velocity function in unit vector notation.
  • Describe the motion of a particle with a constant acceleration in three dimensions.
  • Use the one-dimensional motion equations along perpendicular axes to solve a problem in two or three dimensions with a constant acceleration.
  • Express the acceleration in unit vector notation.

Instantaneous Acceleration

In addition to obtaining the displacement and velocity vectors of an object in motion, we often want to know its acceleration vector at any point in time along its trajectory. This acceleration vector is the instantaneous acceleration and it can be obtained from the derivative with respect to time of the velocity function, as we have seen in a previous chapter. The only difference in two or three dimensions is that these are now vector quantities. Taking the derivative with respect to time v(t),v(t), we find

a(t)=limt0v(t+Δt)v(t)Δt=dv(t)dt.a(t)=limt0v(t+Δt)v(t)Δt=dv(t)dt.
(4.8)

The acceleration in terms of components is

a(t)=dvx(t)dti^+dvy(t)dtj^+dvz(t)dtk^.a(t)=dvx(t)dti^+dvy(t)dtj^+dvz(t)dtk^.
(4.9)

Also, since the velocity is the derivative of the position function, we can write the acceleration in terms of the second derivative of the position function:

a(t)=d2x(t)dt2i^+d2y(t)dt2j^+d2z(t)dt2k^.a(t)=d2x(t)dt2i^+d2y(t)dt2j^+d2z(t)dt2k^.
(4.10)

Example 4.4

Finding an Acceleration Vector A particle has a velocity of v(t)=5.0ti^+t2j^2.0t3k^m/s.v(t)=5.0ti^+t2j^2.0t3k^m/s. (a) What is the acceleration function? (b) What is the acceleration vector at t = 2.0 s? Find its magnitude and direction.

Solution (a) We take the first derivative with respect to time of the velocity function to find the acceleration. The derivative is taken component by component:

a(t)=5.0i^+2.0tj^6.0t2k^m/s2.a(t)=5.0i^+2.0tj^6.0t2k^m/s2.

(b) Evaluating a(2.0s)=5.0i^+4.0j^24.0k^m/s2a(2.0s)=5.0i^+4.0j^24.0k^m/s2 gives us the direction in unit vector notation. The magnitude of the acceleration is |a(2.0s)|=5.02+4.02+(−24.0)2=24.8m/s2.|a(2.0s)|=5.02+4.02+(−24.0)2=24.8m/s2.

Significance In this example we find that acceleration has a time dependence and is changing throughout the motion. Let’s consider a different velocity function for the particle.

Example 4.5

Finding a Particle Acceleration A particle has a position function r(t)=(10tt2)i^+5tj^+5tk^m.r(t)=(10tt2)i^+5tj^+5tk^m. (a) What is the velocity? (b) What is the acceleration? (c) Describe the motion from t = 0 s.

Strategy We can gain some insight into the problem by looking at the position function. It is linear in y and z, so we know the acceleration in these directions is zero when we take the second derivative. Also, note that the position in the x direction is zero for t = 0 s and t = 10 s.

Solution (a) Taking the derivative with respect to time of the position function, we find

v(t)=(102t)i^+5j^+5k^m/s.v(t)=(102t)i^+5j^+5k^m/s.

The velocity function is linear in time in the x direction and is constant in the y and z directions.

(b) Taking the derivative of the velocity function, we find

a(t)=−2i^m/s2.a(t)=−2i^m/s2.

The acceleration vector is a constant in the negative x-direction.

(c) The trajectory of the particle can be seen in Figure 4.9. Let’s look in the y and z directions first. The particle’s position increases steadily as a function of time with a constant velocity in these directions. In the x direction, however, the particle follows a path in positive x until t = 5 s, when it reverses direction. We know this from looking at the velocity function, which becomes zero at this time and negative thereafter. We also know this because the acceleration is negative and constant—meaning, the particle is decelerating, or accelerating in the negative direction. The particle’s position reaches 25 m, where it then reverses direction and begins to accelerate in the negative x direction. The position reaches zero at t = 10 s.

An x y z coordinate system is shown. All the axes show distance in meters and run from -50 to 50 meters. A series of 10 red dots are shown, with the sixth dot is labeled as t = 6 s and the tenth as t = 10 s. The red series of dots starts at the origin and curves upward (both y and z increasing with time). Vertical dashed lines connect the red dots to a series of blue dots in the x y plane. The blue dots are all in the first quadrant (positive x and y). The dots are regularly spaced along the y coordinate, while the x coordinate starts at 0, increases, reaches a maximum of x = 25 m at t = 5, and then decreases back to x = 0 at t 10 s.
Figure 4.9 The particle starts at point (x, y, z) = (0, 0, 0) with position vector r=0.r=0. The projection of the trajectory onto the xy-plane is shown. The values of y and z increase linearly as a function of time, whereas x has a turning point at t = 5 s and 25 m, when it reverses direction. At this point, the x component of the velocity becomes negative. At t = 10 s, the particle is back to 0 m in the x direction.

Significance By graphing the trajectory of the particle, we can better understand its motion, given by the numerical results of the kinematic equations.

Check Your Understanding 4.2

Suppose the acceleration function has the form a(t)=ai^+bj^+ck^m/s2,a(t)=ai^+bj^+ck^m/s2, where a, b, and c are constants. What can be said about the functional form of the velocity function?

Constant Acceleration

Multidimensional motion with constant acceleration can be treated the same way as shown in the previous chapter for one-dimensional motion. Earlier we showed that three-dimensional motion is equivalent to three one-dimensional motions, each along an axis perpendicular to the others. To develop the relevant equations in each direction, let’s consider the two-dimensional problem of a particle moving in the xy plane with constant acceleration, ignoring the z-component for the moment. The acceleration vector is

a=a0xi^+a0yj^.a=a0xi^+a0yj^.

Each component of the motion has a separate set of equations similar to Equation 3.10Equation 3.14 of the previous chapter on one-dimensional motion. We show only the equations for position and velocity in the x- and y-directions. A similar set of kinematic equations could be written for motion in the z-direction:

x(t)=x0+(vx)avgtx(t)=x0+(vx)avgt
(4.11)
vx(t)=v0x+axtvx(t)=v0x+axt
(4.12)
x(t)=x0+v0xt+12axt2x(t)=x0+v0xt+12axt2
(4.13)
vx2(t)=v0x2+2ax(xx0)vx2(t)=v0x2+2ax(xx0)
(4.14)
y(t)=y0+(vy)avgty(t)=y0+(vy)avgt
(4.15)
vy(t)=v0y+aytvy(t)=v0y+ayt
(4.16)
y(t)=y0+v0yt+12ayt2y(t)=y0+v0yt+12ayt2
(4.17)
vy2(t)=v0y2+2ay(yy0).vy2(t)=v0y2+2ay(yy0).
(4.18)

Here the subscript 0 denotes the initial position or velocity. Equation 4.11 to Equation 4.18 can be substituted into Equation 4.2 and Equation 4.5 without the z-component to obtain the position vector and velocity vector as a function of time in two dimensions:

r(t)=x(t)i^+y(t)j^andv(t)=vx(t)i^+vy(t)j^.r(t)=x(t)i^+y(t)j^andv(t)=vx(t)i^+vy(t)j^.

The following example illustrates a practical use of the kinematic equations in two dimensions.

Example 4.6

A Skier Figure 4.10 shows a skier moving with an acceleration of 2.1m/s22.1m/s2 down a slope of 15°15° at t = 0. With the origin of the coordinate system at the front of the lodge, her initial position and velocity are

r(0)=(75.0i^50.0j^)mr(0)=(75.0i^50.0j^)m

and

v(0)=(4.1i^1.1j^)m/s.v(0)=(4.1i^1.1j^)m/s.

(a) What are the x- and y-components of the skier’s position and velocity as functions of time? (b) What are her position and velocity at t = 10.0 s?

An illustration of a skier in an x y coordinate system is shown. The skier is moving along a line that is 15 degrees below the horizontal x direction and has an acceleration of a = 2.1 meters per second squared also directed in his direction of motion. The acceleration is represented as a purple arrow.
Figure 4.10 A skier has an acceleration of 2.1m/s22.1m/s2 down a slope of 15°.15°. The origin of the coordinate system is at the ski lodge.

Strategy Since we are evaluating the components of the motion equations in the x and y directions, we need to find the components of the acceleration and put them into the kinematic equations. The components of the acceleration are found by referring to the coordinate system in Figure 4.10. Then, by inserting the components of the initial position and velocity into the motion equations, we can solve for her position and velocity at a later time t.

Solution (a) The origin of the coordinate system is at the top of the hill with y-axis vertically upward and the x-axis horizontal. By looking at the trajectory of the skier, the x-component of the acceleration is positive and the y-component is negative. Since the angle is 15°15° down the slope, we find

ax=(2.1m/s2)cos(15°)=2.0m/s2ax=(2.1m/s2)cos(15°)=2.0m/s2
ay=(−2.1m/s2)sin15°=−0.54m/s2.ay=(−2.1m/s2)sin15°=−0.54m/s2.

Inserting the initial position and velocity into Equation 4.12 and Equation 4.13 for x, we have

x(t)=75.0m+(4.1m/s)t+12(2.0m/s2)t2x(t)=75.0m+(4.1m/s)t+12(2.0m/s2)t2
vx(t)=4.1m/s+(2.0m/s2)t.vx(t)=4.1m/s+(2.0m/s2)t.

For y, we have

y(t)=−50.0m+(−1.1m/s)t+12(−0.54m/s2)t2y(t)=−50.0m+(−1.1m/s)t+12(−0.54m/s2)t2
vy(t)=−1.1m/s+(−0.54m/s2)t.vy(t)=−1.1m/s+(−0.54m/s2)t.

(b) Now that we have the equations of motion for x and y as functions of time, we can evaluate them at t = 10.0 s:

x(10.0s)=75.0m+(4.1m/s2)(10.0s)+12(2.0m/s2)(10.0s)2=216.0mx(10.0s)=75.0m+(4.1m/s2)(10.0s)+12(2.0m/s2)(10.0s)2=216.0m
vx(10.0s)=4.1m/s+(2.0m/s2)(10.0s)=24.1m/svx(10.0s)=4.1m/s+(2.0m/s2)(10.0s)=24.1m/s
y(10.0s)=−50.0m+(−1.1m/s)(10.0s)+12(−0.54m/s2)(10.0s)2=−88.0my(10.0s)=−50.0m+(−1.1m/s)(10.0s)+12(−0.54m/s2)(10.0s)2=−88.0m
vy(10.0s)=−1.1m/s+(−0.54m/s2)(10.0s)=−6.5m/s.vy(10.0s)=−1.1m/s+(−0.54m/s2)(10.0s)=−6.5m/s.

The position and velocity at t = 10.0 s are, finally,

r(10.0s)=(216.0i^88.0j^)mr(10.0s)=(216.0i^88.0j^)m
v(10.0s)=(24.1i^6.5j^)m/s.v(10.0s)=(24.1i^6.5j^)m/s.

The magnitude of the velocity of the skier at 10.0 s is 25 m/s, which is 60 mi/h.

Significance It is useful to know that, given the initial conditions of position, velocity, and acceleration of an object, we can find the position, velocity, and acceleration at any later time.

With Equation 4.8 through Equation 4.10 we have completed the set of expressions for the position, velocity, and acceleration of an object moving in two or three dimensions. If the trajectories of the objects look something like the “Red Arrows” in the opening picture for the chapter, then the expressions for the position, velocity, and acceleration can be quite complicated. In the sections to follow we examine two special cases of motion in two and three dimensions by looking at projectile motion and circular motion.

Interactive

At this University of Colorado Boulder website, you can explore the position velocity and acceleration of a ladybug with an interactive simulation that allows you to change these parameters.

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